Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 2} = \dfrac{x + 30}{x - 2}$
Multiply both sides by $x - 2$ $ \dfrac{x^2}{x - 2} (x - 2) = \dfrac{x + 30}{x - 2} (x - 2)$ $ x^2 = x + 30$ Subtract $x + 30$ from both sides: $ x^2 - (x + 30) = x + 30 - (x + 30)$ $ x^2 - x - 30 = 0$ Factor the expression: $ (x + 5)(x - 6) = 0$ Therefore $x = -5$ or $x = 6$ The original expression is defined at $x = -5$ and $x = 6$, so there are no extraneous solutions.